The symbol
$$n\choose k$$
is a shorthand for "The number of ways to choose $k$ objects from a collection of $n$ objects."
That is: you have $n$ objects in a line. You select $k$ of them, and paint them red, and you paint the other $n-k$ blue. How many ways are there to do that so that the final result looks different?
A moment of reflection will convince you that you could have chosen $n-k$ of the objects to paint blue first, and painted the rest of the objects red. The answer shouldn't depend on something arbitrary like whether you pick up the blue or the red paint first. If you can grok this fact, then you will have proved to yourself that
$${n\choose k} = {n\choose n-k}$$
How does this relate to binomial expansions? Well, say that we want to expand
$$(r + b)^n$$
Then we'll get a bunch of terms, each of which is a product of some $r$s with some $b$s. There are $n$ brackets, so there will be $n$ multipliers in each of the terms, and each of them will be either $r$ or $b$, so every term looks like
$$a_kr^kb^{n-k}$$
where $a_k$ is a coefficient, and $k=0, 1, 2, \dots n$.
What is $a_k$? Notice that in every bracket, we have to choose either an $r$ or a $b$. We can imagine going through the brackets one by one and making these choices, or you can imagine looking at all the brackets at once, and choosing which of them to pick $r$ form and which to pick $b$ from. The coefficient $a_k$ is the number of different ways of picking $k$ $r$s and $(n-k)$ $bs$ from the $n$ brackets. That is:
$$a_k = {n\choose k}$$
which, using the fact from earlier, we can instantly use to prove that the coefficients in a binomial expansion are symmetric:
$$a_k = {n\choose k} = {n\choose n-k} = a_{n-k}$$
Maybe you can see how binomial expansions relate to Pascal's triangle now?
Hint: It is linked to another cool and exciting relationship between the binomial coefficients!
Sorry if there is a rule against posting on questions that have not been active lately, feel free to delete if there is, but I thought this method might be helpful for someone else who doesn't want to look at Pascal's triangle or won't be given it in a test.
Here is a well-known alternative method you might find faster. Let
$$\binom{n}{k}:=m_k.$$ Then
$$\frac{m_{k+1}}{m_k}=\frac{k!\left(n-k\right)!}{\left(k+1\right)!\left(n-k-1\right)!}=\frac{n-k}{k+1}.$$ This means if you know one coefficient you can calculate all the rest, by multiplying the first coefficient $m_k$ by the corresponding index $n-k,$ then dividing by the number of terms you have in the sum so far, $k+1.$ So for example, if you have $$(a+b)^n=\sum\limits_{k=0}^{n}\binom{n}{k}a^{n-k}b^k,$$ then if we take $n=37,$ you know $(a+b)^{37}=a^{37}+\cdots,$ so your first coefficient is $1.$ It corresponds to the index $n-k=37,$ and you have one term so far. So, your next coefficient is
$$1\times\frac{37}{1}=37.$$ Next, $37\times36/2=666.$ The fourth coefficient is $666\times35/3=7770,$ getting $$(a+b)^{37}=a^{37}+37a^{36}b+666a^{35}b^2+7770a^{34}b^3+\cdots$$and so on until you get half of them and then use the symmetrical nature of the binomial theorem to write down the other half.
Best Answer
Let $\omega=\exp\big(i\frac{\pi}3\big)\cdot$ Observe that, for any integer $k$ we have $\sum_{n=0}^5 \omega^{nk}= \begin{cases} \frac{\omega^{6k-1}}{\omega^k-1}=0&\text{if}\ k\notin6\mathbb{Z},\\[3pt] 6&\text{if}\ k\in6\mathbb{Z}. \end{cases}$
From the Newton binomial formula \begin{eqnarray*} \sum_{n=0}^5(1+\omega^n)^{32} &=&\sum_{n=0}^5\sum_{k=0}^{32}\binom{32}{k}\omega^{nk} =\sum_{k=0}^{32}\binom{32}{k}\sum_{n=0}^5\omega^{nk}\\ &=&\kern-0.7em\sum\limits_{k\in\{0,6,12,18,24,30\}}\kern-0.6em \binom{32}{k}\times6\\ &=&6\sum_{r=0}^5\binom{32}{6r}. \end{eqnarray*}
So the required sum is $S=\big(\sum_{n=0}^5(1+\omega^n)^{32}\big)\big/6$.
Now we have $1+\omega^s =\left\{ \begin{array}{rcl} 2&\text{if}\ s=0,\\[3pt] \sqrt3\exp\big(i\frac\pi6\big)&\text{if}\ s=1,\\[3pt] \exp\big(i\frac\pi3\big)&\text{if}\ s=2,\\[3pt] 0&\text{if}\ s=3,\\[3pt] \exp\big({-}i\frac\pi3\big)&\text{if}\ s=4,\\[3pt] \sqrt3\exp\big({-}i\frac\pi6\big)&\text{if}\ s=5. \end{array} \right.$
In the end we obtain \begin{eqnarray*} S&{}={}&\bigg[2^{32} +3^{16}\!\exp\Big({-}i\frac{2\pi}3\Big) +\exp\Big(i\frac{2\pi}3\Big) +0 +\exp\Big({-}i\frac{2\pi}3\Big) +3^{16}\!\exp\Big(i\frac{2\pi}3\Big)\bigg]\mathbin{\Big/}6\\ &{}={}&\frac{2^{32}-3^{16}-1}6\\ &{}={}&708653429. \end{eqnarray*}
Edit (to answer a comment below).
I don't see any valuable generalization of the method above. Suppose that one replaces 32 by $m$, 6 by $\lambda$ and puts $\ell=\lfloor m\lambda\rfloor$, in order to calculate $\sum_{n=0}^\ell\binom{m}{\lambda n}$. We would take $\omega=\exp(2i\pi/\lambda)$, but the complication is that there's no reason in general to have $\sum\limits_{n\leqslant\ell}\omega^{kn}=0$ for $k\notin\lambda\mathbb{Z}$, as above. In general this sum would be $\frac{\omega^{k(\ell+1)}-1}{\omega^k-1}\cdot$
With the values $(m,\lambda)=(32,6)$ given in the problem, everything went fine because $\ell+1\in\lambda\mathbb{Z}$ (i. e. $5+1\in6\mathbb{Z}$.
Perhaps the general $\displaystyle\sum_{n=0}^\ell\binom{m}{\lambda n}$ can be calculated by other means?